Let $G$ be a group acting by homeomorphisms on the standard Cantor set $X$ (or if you prefer, $G$ is acting on the countable atomless Boolean algebra, which can be realised as the clopen subsets of $X$). Say a non-empty clopen subset $\alpha$ of $X$ is minorising (for $G$) if the following holds:

For every non-empty clopen subset $\beta$ of $X$, there is some $g \in G$ such that $g\beta$ contains $\alpha$.

Now suppose that there is a minorising subset $\alpha$, and that $G\alpha$ (that is the set of $g\alpha$ for $g \in G$) covers $X$. Let $n$ be the smallest size of a subcover of $G\alpha$, that is the smallest number of $G$-translates of $\alpha$ needed to cover $X$. It is clear that $n$ is finite (by compactness) and at least $2$, and that $n$ does not depend on the choice of $\alpha$ (beyond ensuring that $\alpha$ is minorising), so it is an invariant of the action.

The condition $n=2$ is equivalent to the condition that every proper non-empty clopen subset is minorising. (This occurs for instance if $G$ consists of every homeomorphism of $X$.)

Questions: Can $n$ be greater than $2$? What if $G$ is simple, and/or every orbit of $G$ on $X$ is dense?

If $n=2$, can $G$ still have infinitely many orbits on the clopen subsets of $X$?

NB: The existence of a minorising subset means that $G$ destroys most of the extra structure I can think of putting on $X$, such as non-trivial metrics and measures. (Can $G$ act by quasi-isometries?) Effectively $\alpha$ is `as small as possible' up to the action of $G$.

If $\alpha$ is minorising (which by the way is equivalent to saying that $G\alpha$ is a $\pi$-base for $X$) and $G\alpha$ covers $X$ then necessarily every orbit of $G$ on $X$ is dense.
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Ramiro de la VegaDec 16 '11 at 16:34

Ah yes, good point. This is a space in which every nonempty open set contains a nonempty clopen set.
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Colin ReidDec 16 '11 at 17:21

2 Answers
2

The answer to your last question is YES. Maybe there is a simpler way to see this, but here is a way:

If $G$ acts freely on $X$ and $\alpha$ is minorising (with $n=2$ if you wish), then there is a zero-dimensional compact space $Z$ of uncountable weight, a suryective map $\pi:Z \to X$ and an action of $G$ on $Z$ commuting with $\pi$ such that $\pi^{-1}\alpha$ is minorising. If $G$ was countable then necessarily $G$ has infinitely many (in fact uncountably many) orbits on the clopen subsets of $Z$. Now you can just consider a suitable countable subalgebra of clopens of $Z$ for which you still have infinitely many orbits and a minorising element (still with $n=2$ if you wish).

Edit: The answer to the first question is also YES. If $G$ acts on $X$ then $G \times \mathbb{Z}_2$ acts in a natural way on $Z$ the disjoint union of two Cantor spaces (which is again a Cantor space), so that the $n$ of a minorising subset $\alpha$ of $X$ gets doubled when $\alpha$ is viewed as a subset of one of the copies in $Z$.
Unfortunately $G \times \mathbb{Z}_2$ is not simple. Also, I don´t know if one can get $n=3$ (or any odd number).

Could you explain a bit more your construction of $\pi$? Why is $\pi^{-1}\alpha$ minorising?
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Jesse PetersonDec 16 '11 at 23:43

Jesse: You can see a detailed construction in my article "Basic homogeneity in the class of zero-dimensional compact spaces". The construction uses $CH$ because there I wanted to make sure that the group of autohomeomorphisms of $Z$ was generated by G (i.e. any autohomeomorphism of $Z$ is obtained by pasting finitely many elements of G). Since we don't want that here, we can avoid using $CH$. Also, in the construction, sets of the form $\pi^{-1}U$ with $U$ clopen in $X$ form a $\pi$-base for $Z$ and that is why $\pi^{-1}\alpha$ is still minorising.
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Ramiro de la VegaDec 17 '11 at 13:06

@Ramiro: Perhaps there is some language or notation that I do not understand, but if $\pi: Z \to X$ is not injective then I don't see why sets of the form $\pi^{-1}U$ should separate points in $Z$.
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Jesse PetersonDec 17 '11 at 22:30

@Jesse: They don't separate points, they just form a $\pi$-base; this just means that any open subset of $Z$ contains a set of the form $\pi^{-1}U$.
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Ramiro de la VegaDec 18 '11 at 0:52

I ran across this question after pondering your more recent question here.

Ramiro has already described examples giving a YES answer to your final question, but here is a large and naturally arising class of examples.

Take $G$ to be the full automorphism group $Aut(T)$ of any finite valence tree $T$ on which $Aut(T)$ acts cocompactly, and take $X$ to be the Cantor set of ends of $T$. Or, you can take $G$ to be any cocompact subgroup of $Aut(T)$, such as the free group of finite rank $n \ge 2$ acting on the universal cover of a rose with $n$ petals, or the fundamental group of any finite graph of groups with finite vertex and edge groups, acting on its Bass-Serre tree.

For such actions, every clopen is minorizing and your number $n$ always equals $2$. The proof uses the fact every orbit of the action of $G$ on $X$ is dense, and that the set of source-sink pairs $x \ne y \in X$ is dense in $X \times X$; these are pairs for which there exists $g \in G$ whose action on $X$ has source--sink dynamics with source $x$ and sink $y$. So, if $C,D$ are any proper clopens then choose a source-sink pair $x,y$ with $x \in D$, $y \not\in D$, then choose a translate of $C$ that contains $x$, then choose $n$ so large that $g^n(C)$ contains $D$. Similarly, if $C$ is any proper clopen in $X$, choose a translate $C_1$ that contains $x$, choose another translate $C_2$ that contains $y$, then choose sufficiently large $n$ so that the pair $g^n(C_1),C_2$ covers $X$.

Also for such actions, there are infinitely many $G$-orbits of clopens. To see this, given any clopen $C$, and letting $D=X-C$, the intersection in $T$ of the convex hulls of $C$ and $D$ is a finite subtree of $T$, the number of vertices of this subtree is an invariant of the orbit of $C$, and this number can be as large as you like.

Indeed, I was having similar thoughts just a few days ago. There are also plenty of simple groups that are of this kind, e.g. they appear in a paper of Burger and Mozes (don't have the reference to hand at the moment). Getting n>2 for a simple group seems harder, though.
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Colin ReidMay 21 '12 at 12:54